The Helmholtz free energy UF plays an important role for systems where T, U and V are fixed

- F is minimum in equilibrium, when U,V and T are fixed!

dddUdF by using: pdVdUd pdVddF

immediate relations:

pV

F

F

V

V

pV

Maxwell relation:

VV

Up

Calculation of the Helmholtz free energy (F) from the partition function (Z):

(proof: by showing that F=-ln(Z) satisfy the F=U- =U+(F/)V relation, or by the use of the Renyi entropy formula --> see extra problem)

)ln(ZF

Immediate relations:

/exp)exp(

)(

)exp(

s

s

s FZ

P

FZ

Ideal gas: A first look. One atom in a box

We first calculate the partition function (Z1) of one atom of mass M free to move in a cubical box of volume V=L3

)/sin()/sin()/sin(),,( LxnLxnLxnAzyx zyx Wave functions of possible states:

Energies of possible states: )(2

22222

zyxn nnnLM

(nx , ny , nz : positive integers)

)](exp[]2/)(exp[ 222

0 0

2

0

22222

}{ }{ }{

21 zyxzyxzyx

n n nnnndndndnMLnnnZ

x y z

where2222 2/ ML after performing the integrals: 2/32

32/31 )/2(

8/

M

VZ

We can introduce the so called quantum concentration, which is rouhgly the concentration of one atom in a cube of side equal to the thermal average de Broglie wavelength. ( )

2/32 )2/( MnQ 2/1)/(/ MvM

n

nVnZ Q

Q 1Whenever n<<nQ --> classical regime. An ideal gas is defined as a gas

of nonintearcting atoms in the classical regime!

TkZZ

U Bn

nn

2

3

2

3)/)ln((

)/exp(1

2

11

Average internal energy of one particle

Thermal average occupancy of one state:

(for the classical regime, this must be <<1!)

Qn nnZZ /)/exp( 11

11

N atoms in a box

If we have N non-interacting, independent and distinguishable particles in a box:

}{}{}{},...,{}{

)/exp(...)/exp()/exp(]/)...(exp[)/exp(2

21

12121 N

NNN s

ss

ss

ssssssss

sZ

NZZZZ ...21

if the particles are identical and indistinguishableNZ

NZ 1!

1

For an ideal gas composed of N molecules we have NQVn

NZ )(

!

1

total energy of an ideal gas: NkTNM

V

NZU

N

2

3

2

3)/2(!

1ln

)/)ln((2/32

22

thermal equation of state of ideal gases: VNkTVNV

M

V

N

V

Fp

N

//)/2(!

1ln

2/32

entropy of an ideal gas: 2

5)/[ln(

)/2(!

1ln

2/32

nnNM

V

NFQ

N

V

Sackur-Tetrode equation

Problems

1. Problem nr. 1 (Free energy of a two state system) on page 81

2. Problem nr. 2 (Magnetic susceptibility) on page 81

3. Problem nr. 3 (Free energy of a harmonic oscillator) on page 82

4. Problem nr. 4 (Energy fluctuations) on page 83

Extra problem

Consider a closed thermodynamic system (N constant) with fixed temperature (T) and volume

(V). By using the Renyi entropy formula, the expression for the probability of one state, and the

fact that F=U-TS, prove, that F=-kBTln(Z)

Equipartition of energy for ideal gases

- in an ideal gas for all possible degrees of freedom the average thermal energy is: /2 (kT/2)

- generalization: whenever the hamiltonian of the system is homogeneous of degree 2 in a canonical momentum component, the classical limit of the thermal average kinetic energy associated with that momentum will be /2

degrees of freedom for one molecule:

- molecules composed by one atom: 3 --> motion in the three direction of the space

- molecules composed by two atom: 7 --> motion of the molecule in the three directions of space + rotations around the two axis perpendicular to the line connecting the two atoms + vibrations (kinetic and potential energy for this)

degrees of freedom for the system: N x degrees of freedom for one molecule

Heat capacity at constant volume of one molecule of H2

in the gas phase